Right-inverse of trace extension operator

Question

Let $\Omega\subset\mathbb{R}^n$ be bounded with Lipschitz boundary. Then we know that there exists a bounded linear trace operator $T:W^{1,1}(\Omega)\to L^1(\partial \Omega)$. Howerever I want to know does this operator have the right inverse operator $E$ with $E:L^1(\partial\Omega)\to W^{1,1}(\Omega)$ such that $E$ is linear and bounded.

Answer 1

This is true, and it is a result of Gagliardo. Analogously to the case of general $p,$ the idea is to prove it in the half space by defining an extension operator of the form $$ Tf(x) = \int_{\Bbb R^{n-1}} \rho(t) f(x'+x_n t) \,\mathrm{d}t $$ where we write $x=(x',x_n) \in \Bbb R^{n-1} \times \Bbb R,$ and $\rho$ is a standard radial mollifier in $\Bbb R^{n-1}.$ One can check that if $f \in L^1(\Bbb R^{n-1}),$ then $\nabla (Tf) \in L^1(\Bbb R^n_+)$ defining a right inverse $L^1(\Bbb R^{n-1}) \to \dot{W}^{1,1}(\Bbb R^n_+).$ The general case follows by a patching argument; a detailed proof can be found in the following text:

Leoni, Giovanni, A first course in Sobolev spaces, Graduate Studies in Mathematics 181. Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-2921-8/hbk; 978-1-4704-4226-2/ebook). xxii, 734 p. (2017). ZBL1382.46001.

Namely it is the content of Theorem 18.18 in the 2nd edition. Note the text uses a simpler extension due to Mironescu, but the above is commonly seen as it works in other scales also.